Question

▼ 6:01 ← PHYS-1101-02 Lect 50.7 m/s LEARN MORE REMARKS The trooper, traveling about twice as fast as the car, must swerve or apply his brakes strongly to avoid a collision! This problem can also be solved graphically by plotting position versus time for each vehicle on the same graph The intersection of the two graphs corresponds to the time and position at which the trooper overtakes the car QUESTION The graphical solution corresponds to finding the intersection of what two types of curves in the txplane? Select all that apply) a straight horizontal line a straight line sloped upward an upward-shaped curve whose slope increases as the displacement x increases a downward-shaped curve whose slope decreases and then increases as x increases a straight line sloped downward PRACTICE IT Use the worked example above to help you solve this problem. A car traveling at a constant speed of 28.6 m/s passes a trooper hidden behind a bilboard, as in the figure. One second after the speeding car passes the billboard, the trooper sets off in chase with a constant acceleration of 3.57 m/s (a) How long does it take the trooper to overtake the speeding car? (b) How fast is the trooper going at that time? m/s EXERCISE HINTS: GETTING STARTED I IMSTUCK A motorist with an expired license tag travels at a constant speed of 19.1 m/s down a street, and a policeman on a motorcycle, taking another 4.18 s to finish his donut, gives chase at an acceleration of 2.62 m/s2 (a) Find the time required to catch the car (b) Find the distance the trooper travels while overtaking the motorist Need Help? Read It Submit Save Assignment Progress 83 Dashboard

Please help me out as much as possible, I'm stuck, please try and get to the ones you can(: Thanks bunches

Answer 1

a)

consider the motion of the policeman :

t = time of travel to catch the motorist

X_p = displacement of policeman

a_p = acceleration = 2.62 m/s²

v_o = initial velocity = 0 m/s

using the equation

X_p = v_o t + (0.5) a_p t²

X_p = (0) t + (0.5) (2.62) t²

X_p = (1.31) t² eq-1

consider the motion of the motorist :

t_m = time of travel for the motorist = t + 4.18

X_m = displacement of motorist

v_m = constant velocity of motorist = 19.1 m/s

using the equation

X_m = v_m t_m

X_m = 19.1 (t + 4.18)      eq-2

for the catch to be successful

X_p = X_m

(1.31) t² =  19.1 (t + 4.18)

t = 17.97 sec

b)

using eq-1

X_p = (1.31) t²

X_p = (1.31) (17.97)²

X_p = 423 m